What is the derivative of sqrt(a+bx) ?

The derivative of sqrt(a+bx) is b/(2sqrt(a+bx))

What is the first derivative of sqrt(a+bx) ?

The first derivative of sqrt(a+bx) is b/(2sqrt(a+bx))

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\frac{d}{d x} (a + b x)

\frac{b}{2 a + b x}

Solution steps

\frac{d}{d x} (a + b x)

Apply the chain rule:

\frac{1}{2 a + b x} \frac{d}{d x} (a + b x)

= \frac{1}{2 a + b x} \frac{d}{d x} (a + b x)

\frac{d}{d x} (a + b x) = b

= \frac{1}{2 a + b x} b

Simplify

\frac{1}{2 a + b x} b : \frac{b}{2 a + b x}

= \frac{b}{2 a + b x}

\frac{d}{d x} (\frac{2 x e ^{x}}{x ^{2} + 1})

\int 7 x^{2} sin (2 x) d x

k = 1 \sum \infty (\frac{1}{2})^{k}

\frac{d}{d x} (\frac{x ^{2}}{64} + \frac{y ^{2}}{25}) = 1

x \to 2 - lim (\frac{- 1}{x ^{2} - 4})

What is the derivative of sqrt(a+bx) ?
The derivative of sqrt(a+bx) is b/(2sqrt(a+bx))
What is the first derivative of sqrt(a+bx) ?
The first derivative of sqrt(a+bx) is b/(2sqrt(a+bx))